Line

2021-01-05

A population of values has a normal distribution with $\mu =99.6$ and $\sigma =35.1$. You intend to draw a random sample of size $n=84$.
Find the probability that a sample of size $n=84$ is randomly selected with a mean between 98.5 and 100.7.
$P\left(98.5<\stackrel{―}{X}<100.7\right)=$?

saiyansruleA

Step 1
From the provided information,
Mean $\left(\mu \right)=99.6$
Standard deviation $\left(\sigma \right)=35.1$
$X\sim N\left(99.6,35.1\right)$
Sample size $\left(n\right)=84$
Step 2
The required probability that a sample of size $n=84$ is randomly selected with a mean between 98.5 and 100.7 can be obtained as:
and 100.7 can be obtained as:
$P\left(98.5<\stackrel{―}{X}<100.7\right)=P\left(\frac{98.5-99.6}{\frac{35.1}{\sqrt{84}}}<\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{100.7-99.6}{\frac{35.1}{\sqrt{84}}}\right)$
$=P\left(-0.287
$=P\left(Z<0.287\right)-P\left(Z<-0.287\right)$
$=P\left(Z<0.287\right)-\left[1-P\left(z<0.287\right)\right]$
$=2P\left(Z<0.287\right)-1$
$=2\left(0.6129\right)-1=0.2258$ (Using standard normal table)
Thus, the required probability is 0.2258.

Do you have a similar question?